AtCoder Beginner Contest 182----E. Akari

旧人旧事更值得怀念，或悲或喜每个人身上都承载着某一段时光，但是我会永远清澈永远疯狂永远浪漫。没有人能回到过去重新活过，但你我都可以从现在开始，决定我们未来的模样。
                                                                                                                                                                           ----喻言

 

 题解：将所有灯和墙都放到矩形中，然后逐行从左到右扫描一遍，再从右到左扫描一遍；逐列从上到下扫描一遍，再从下到上扫描一遍。最后统计亮着的格子即可（也就是大于0的格子数）。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <ctime>
#include <cctype>
#include <bitset>
#include <utility>
#include <complex>
#include <iomanip>
#include <numeric>
#include<unordered_set>
#include <climits>//INT_100010n
#include <regex>
//#include<bits/stdc++.h>
#define PP pair<ll,int>
#define inf 0x3f3f3f3f
#define INF 0x7fffffff;
#define llinf 0x3f3f3f3f3f3f3f3fll
#define dinf 1000000000000.0
#define PI 3.1415926
#define LL unsigned int
#define mod 1000000007
#define wc 1e-18
typedef long long ll;
using namespace std;
int h, w, n, m;
int x,y,jg;
int mp[1510][1510];
int main()
{
    cin >> h >> w >> n >> m;
    for (int i = 0; i < n; i++)
    {
        cin >> x >> y;
        mp[x - 1][y - 1] = 1;
    }
    for (int i = 0; i < m; i++)
    {
        cin >> x >> y;
        mp[x - 1][y - 1] = -1;
    }
    for (int i = 0; i < h; i++)
    {
        bool fg = false;
        for (int j = 0; j < w; j++)
        {
            if (mp[i][j] == 1)
                fg = true;
            else if (mp[i][j] == -1)
                fg = false;
            else if (fg)
                mp[i][j] = 2;
        }
        fg = false;
        for (int j = w - 1; j >= 0; --j)
        {
            if (mp[i][j] == 1)
                fg = true;
            else if (mp[i][j] == -1)
                fg = false;
            else if (fg)
                mp[i][j] = 2;
        }
    }
    for (int j = 0; j < w; j++)
    {
        bool fg = false;
        for (int i = 0; i < h; i++)
        {
            if (mp[i][j] == 1)
                fg = true;
            else if (mp[i][j] == -1)
                fg = false;
            else if (fg)
                mp[i][j] = 2;
        }
        fg = false;
        for (int i = h - 1; i >= 0; --i)
        {
            if (mp[i][j] == 1)
                fg = true;
            else if (mp[i][j] == -1)
                fg = false;
            else if (fg)
                mp[i][j] = 2;
        }
    }
    for (int i = 0; i < h; i++)
        for (int j = 0; j < w; j++)
            if(mp[i][j]>0)
                jg++;
    cout << jg << endl;
}